Tree
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 20098 | Accepted: 6608 |
Description
Give a tree with n vertices,each edge has a length(positive integer less than 1001).
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Define dist(u,v)=The min distance between node u and v.
Give an integer k,for every pair (u,v) of vertices is called valid if and only if dist(u,v) not exceed k.
Write a program that will count how many pairs which are valid for a given tree.
Input
The input contains several test cases. The first line of each test case contains two integers n, k. (n<=10000) The following n-1 lines each contains three integers u,v,l, which means there is an edge between node u and v of length l.
The last test case is followed by two zeros.
The last test case is followed by two zeros.
Output
For each test case output the answer on a single line.
题意：给一颗带权树，求树上长度不超过L的路径条数
首先有一个有树高限制时的树形DP做法..........
对于一条树路径 只有经过或不经过一个点的情况
考虑经过一个点的路径，可以由其他点到它的两条路径拼出来
对于不经过的情况 把一棵树按这个点拆成好几棵分治
每次对于当前子树选择树的重心，最多递归logn次，而每层最多只有n个点(每层的所有子树组成整棵树)，复杂度O(logn*处理每层的复杂度)
过程：
1.求重心
2.处理经过当前点的路径
3.对子树分治
每次分治的各个子树是互不影响的，vis[i]表示i这个点已经分治过了
注意：既然你的写法是先找重心在递归，那么一定要rt=0;dfsRt(v,0);dfsSol(rt);是rt啊啊啊啊啊不是v了
对于本题，处理经过点u的路径时，先dfs子树中所有点对深度，排序两个指针往里扫计算<=L的，在减去在同一颗子树里的(同样计算)
总复杂度O(nlog^2n)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=,INF=1e9+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x*f;
}
int n,L,u,v,w;
struct edge{
int v,w,ne;
}e[N<<];
int h[N],cnt;
inline void ins(int u,int v,int w){
cnt++;
e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
cnt++;
e[cnt].v=u;e[cnt].w=w;e[cnt].ne=h[v];h[v]=cnt;
} int size[N],d[N],vis[N],root,sum;
void dfsRoot(int u,int fa){
size[u]=;d[u]=;
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]||v==fa) continue;
dfsRoot(v,u);
size[u]+=size[v];
d[u]=max(d[u],size[v]);
}
d[u]=max(d[u],sum-size[u]);
if(d[u]<d[root]) root=u;
}
int deep[N],a[N];
void dfsDeep(int u,int fa){
a[++a[]]=deep[u];
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]||v==fa) continue;
deep[v]=deep[u]+e[i].w;
dfsDeep(v,u);
}
} int cal(int u,int now){
deep[u]=now;a[]=;
dfsDeep(u,);
sort(a+,a++a[]);
int l=,r=a[],ans=;
while(l<r){
if(a[l]+a[r]<=L) ans+=r-l,l++;
else r--;
}
return ans;
}
int ans;
void dfsSol(int u){//printf("dfs %d\n",u);
vis[u]=;
ans+=cal(u,);
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]) continue;
ans-=cal(v,e[i].w);
sum=size[v];
root=;dfsRoot(v,);
dfsSol(root);
}
} int main(){
//freopen("in.txt","r",stdin);
while(true){
n=read();L=read();if(n==) break;
cnt=;memset(h,,sizeof(h));
memset(vis,,sizeof(vis));
ans=;
for(int i=;i<=n-;i++) u=read(),v=read(),w=read(),ins(u,v,w);
sum=n;
root=;d[]=INF;
dfsRoot(,);
dfsSol(root);
printf("%d\n",ans);
}
}
还有一种做法，用treap维护，考虑经过每个点的路径时，建一颗treap维护长度，每个点加上之前遍历过的这点的子树中<L-deep[v]+1的，最后再把这棵子树的所有深度加入treap
//
// main.cpp
// treap
//
// Created by Candy on 2017/1/9.
// Copyright ? 2017年 Candy. All rights reserved.
// #include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define lc t[x].l
#define rc t[x].r
const int N=1e5+,INF=1e9;
int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-; c=getchar();}
while(c>=''&&c<=''){x=x*+c-''; c=getchar();}
return x*f;
} struct node{
int l,r,v,w,size,rnd;
}t[N];
int sz,root;
inline void update(int x){t[x].size=t[lc].size+t[rc].size+t[x].w;}
inline void lturn(int &x){
int c=rc;rc=t[c].l;t[c].l=x;
t[c].size=t[x].size;update(x);x=c;
}
inline void rturn(int &x){
int c=lc;lc=t[c].r;t[c].r=x;
t[c].size=t[x].size;update(x);x=c;
}
void ins(int &x,int v){
if(x==){
x=++sz;
t[x].l=t[x].r=;
t[x].v=v;t[x].w=t[x].size=;
t[x].rnd=rand();
return;
}
t[x].size++;
if(v==t[x].v) t[x].w++;
else if(v<t[x].v){
ins(lc,v);
if(t[lc].rnd<t[x].rnd) rturn(x);
}else{
ins(rc,v);
if(t[rc].rnd<t[x].rnd) lturn(x);
}
}
int que(int x,int v){//cnt of <v
if(!x) return ;
if(t[x].v==v) return t[lc].size;
if(v<t[x].v) return que(lc,v);
else return t[lc].size+t[x].w+que(rc,v);
} int n,L,u,v,w;
struct edge{
int v,w,ne;
}e[N<<];
int h[N],cnt;
inline void ins(int u,int v,int w){
cnt++;
e[cnt].v=v;e[cnt].w=w;e[cnt].ne=h[u];h[u]=cnt;
cnt++;
e[cnt].v=u;e[cnt].w=w;e[cnt].ne=h[v];h[v]=cnt;
}
int vis[N],size[N],f[N],sum,rt;
void dfsRoot(int u,int fa){
size[u]=;f[u]=;
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]||v==fa) continue;
dfsRoot(v,u);
size[u]+=size[v];
f[u]=max(f[u],size[v]);
}
f[u]=max(f[u],sum-size[u]);
if(f[u]<f[rt]) rt=u;
} int ans,deep[N];
void dfsDeep(int u,int fa,int p){
if(p==) ans+=que(root,L-deep[u]+);
else ins(root,deep[u]);
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]||v==fa) continue;
deep[v]=deep[u]+e[i].w;
dfsDeep(v,u,p);
}
}
void dfsSol(int u){//printf("sol %d\n",u);
vis[u]=;
sz=root=;
ins(root,);
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]) continue;
deep[v]=e[i].w;
dfsDeep(v,u,);
dfsDeep(v,u,);
}
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[v]) continue;
sum=size[v];
rt=;dfsRoot(v,u);
dfsSol(rt);
}
}
int main(){
//freopen("in.txt","r",stdin);
while(true){
n=read();L=read();if(n==) break;
cnt=;memset(h,,sizeof(h));
memset(vis,,sizeof(vis));
ans=;
for(int i=;i<=n-;i++) u=read(),v=read(),w=read(),ins(u,v,w);
sum=n;
rt=;f[]=INF;
dfsRoot(,);
dfsSol(rt);
printf("%d\n",ans);
}
}